Question

Obtain the expectation value of the potential energy 2 2
2
1
V (x) = mw x of the onedimensional
harmonic oscillator in the first excited state
/ 2
1/ 2
1
2 2
2
( ) 2 a x ax e
a
x −




p
y =

Answer 1

Answer on Question #51443, Physics, Solid State Physics

Obtain the expectation value of the potential energy $V(x) = \frac{m\omega^2x^2}{2}$ of the one-dimensional harmonic oscillator in the first excited state $\psi_{1}(x) = \frac{a^{3 / 2}\sqrt{2}}{\pi^{1 / 4}} x\exp \left[-\frac{a^{2}x^{2}}{2}\right]$

Solution:

The expectation value of the potential energy $V(x) = \frac{m\omega^2x^2}{2}$ of the one dimensional harmonic oscillator in the first excited state is given by Eq.(1)

$$\left\langle \psi_ {1} \right| V \left| \psi_ {1} \right\rangle = \int_ {- \infty} ^ {+ \infty} \psi_ {1} (x) V (x) \psi_ {1} (x) d x$$

So,

$$\begin{array}{l} \left\langle \psi_ {1} \right| V \left| \psi_ {1} \right\rangle = \int_ {- \infty} ^ {+ \infty} \frac {a ^ {3 / 2} \sqrt {2}}{\pi^ {1 / 4}} x \exp \left[ - \frac {a ^ {2} x ^ {2}}{2} \right] \frac {m \omega^ {2} x ^ {2}}{2} \frac {a ^ {3 / 2} \sqrt {2}}{\pi^ {1 / 4}} x \exp \left[ - \frac {a ^ {2} x ^ {2}}{2} \right] d x = \frac {2 a ^ {3}}{\sqrt {\pi}} \int_ {- \infty} ^ {+ \infty} \frac {m \omega^ {2} x ^ {4}}{2} \exp \left[ - a ^ {2} x ^ {2} \right] d x = \\ \frac {m \omega^ {2}}{a ^ {2} \sqrt {\pi}} \int_ {- \infty} ^ {+ \infty} (a x) ^ {4} \exp \left[ - a ^ {2} x ^ {2} \right] d (a x) = \left| \begin{array}{l} y = (a x) ^ {2} \\ d y = 2 a x d (a x) = 2 \sqrt {y} d (a x) \end{array} \right| = \frac {m \omega^ {2}}{a ^ {2} \sqrt {\pi}} \int_ {- \infty} ^ {+ \infty} y ^ {2} \exp \left[ - y \right] \frac {d y}{2 \sqrt {y}} = \\ = \frac {m \omega^ {2}}{a ^ {2} \sqrt {\pi}} \int_ {- \infty} ^ {+ \infty} y ^ {2} \exp [ - y ] \frac {d y}{2 \sqrt {y}} = \frac {m \omega^ {2}}{2 a ^ {2} \sqrt {\pi}} \int_ {- \infty} ^ {+ \infty} y ^ {3 / 2} \exp [ - y ] d y = \frac {m \omega^ {2}}{2 a ^ {2} \sqrt {\pi}} \Gamma (5 / 2) = \\ \frac {m \omega^ {2}}{2 a ^ {2} \sqrt {\pi}} \cdot \frac {3}{2} \cdot \frac {1}{2} \cdot \Gamma (1 / 2) = \frac {3 m \omega^ {2}}{8 a ^ {2} \sqrt {\pi}} \cdot \sqrt {\pi} = \frac {3 m \omega^ {2}}{8 a ^ {2}} \\ \end{array}$$

where $\Gamma (\xi)$ is the Gamma-function.

So,

$$\left\langle \psi_ {1} \right| V \left| \psi_ {1} \right\rangle = \frac {3 m \omega^ {2}}{8 a ^ {2}}$$

Answer: $\left\langle \psi_{1} \right| V \left| \psi_{1} \right\rangle = \frac{3m\omega^{2}}{8a^{2}}$